Which is the luminous object?

(a) Calculate the magnitude of the gravitational force exerted by Mars on a 80 kg human standing on the surface of Mars. (The ma

andrew11 [14]

Answer:

a) $F=1.044\times 10^9\ N$

b) $F'=1.044\times 10^9\ N$

c) $F_p=1.0672\times10^{-7}\ N$

d) Treat the humans as though they were points or uniform-density spheres.

Explanation:

Given:

mass of Mars, $M=6.4\times 10^{23}\ kg$

radius of the Mars, $r=3.4\times 10^{6}\ m$

mass of human, $m=80\ kg$

a)

Gravitation force exerted by the Mars on the human body:

$F=G.\frac{M.m}{r^2}$

where:

$G=6.67 \times 10^{-11}\ m^3.kg^{-1}.s^{-2}$ = gravitational constant

$F=6.67\times10^{-11}\times \frac{6.4\times 10^{23}\times 80}{(3.4\times 10^{6})^2}$

$F=1.044\times 10^9\ N$

b)

The magnitude of the gravitational force exerted by the human on Mars is equal to the force by the Mars on human.

$F'=F$

$F'=1.044\times 10^9\ N$

c)

When a similar person of the same mass is standing at a distance of 4 meters:

$F_p=6.67\times10^{-11}\times \frac{80\times 80}{4}$

$F_p=1.0672\times10^{-7}\ N$

d)

The gravitational constant is a universal value and it remains constant in the Universe and does not depends on the size of the mass.

Yes, we have to treat Mars as spherically symmetric so that its center of mass is at its geometric center.

Yes, we also have to ignore the effect of sun, but as asked in the question we have to calculate the gravitational force only due to one body on another specific body which does not brings sun into picture of the consideration.

4 0

2 years ago

A car covers 400 km in an hour towards west .calculate the velocity

crimeas [40]

Answer:

-400km/hr

Explanation:

Velocity=displacement/time

=400/1

=400Km/hr

=-400km/hr (because west direction)

7 0

2 years ago

A very long solid insulating cylinder has radius R = 0.1 m and uniform charge density rho0= 10-3 C/m3. Find the electric field a

Galina-37 [17]

Answer:

$E = (0.56 \times 10^8 ) r \ \ N/c$

Explanation:

Given that:

$\rho_o = (10^{-3} ) \ c/m^3$

R = (0.1) m

To find the electric field for r < R by using Gauss Law

${\oint}E^{\to}* da^{\to} = \dfrac{Q_{enclosed}}{\varepsilon_o} --- (1)$

For r < R

$Q_{enclosed}=(\rho) ( \pi r^2 ) l$

$E*(2 \pi rl)= \dfrac{\rho ( \pi r ^2 l)}{\varepsilon_o}$

$E= \dfrac{\rho ( r)}{2 \varepsilon_o}$

where;

$\varepsilon_o = 8.85 \times 10^{-12}$

$E= \dfrac{10^{-3} ( r)}{2 (8.85 \times 10^{-12})}$

$E = (0.56 \times 10^8 ) r \ \ N/c$

4 0

2 years ago

Which statement correctly describes the current in a circuit that is made up of any two resistors connected in parallel with a b

AysviL [449]

B is the answer I think

8 0

2 years ago

A silver bar 0.125 meter long is subjected to a temperature change from 200°C to 100°C. What will be the length of the bar after

Delicious77 [7]

$\Delta L= \alpha L_0 (T_f-T_i)$

= (18 x 10^-6 /°C)(0.125 m)(100° C - 200 °C)

= -0.00225 m

New length = L + ΔL
= 1.25 m + (-0.00225 m)
= 1.248

D

5 0

3 years ago

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